The molar heat of fusion of lead (Pb) is 1.140 kcal/mol. How much heat must be added to 97.0 g of solid lead at its melting point in order for it to completely melt?
A 1.44 kJ
B 10.18 kJ
C 12.48 kJ
D 25.04 kJ
q = mass x heat fusion = ?
Remember to change 97.0 g to mols.
To solve this problem, we can use the equation:
q = m * ΔH
Where:
q is the heat (in kilojoules)
m is the mass (in grams)
ΔH is the molar heat of fusion (in kilocalories/mol)
First, let's convert the molar heat of fusion from kilocalories/mol to kilojoules/mol. Since there are 4.184 kilojoules in 1 kilocalorie, we can multiply the given value by 4.184:
ΔH = 1.140 kcal/mol * 4.184 kJ/kcal
ΔH = 4.762 kJ/mol
Next, we need to convert the given mass of 97.0 grams of lead to moles. To do this, we divide the mass by the molar mass of lead. The molar mass of lead (Pb) is approximately 207.2 g/mol:
moles = 97.0 g / 207.2 g/mol
moles = 0.468 mol
Now, we can calculate the heat required to melt the given mass of lead by multiplying the number of moles by the molar heat of fusion:
q = 0.468 mol * 4.762 kJ/mol
q = 2.231 kJ
Therefore, the amount of heat required to completely melt 97.0 g of solid lead is approximately 2.231 kJ. Since none of the answer choices match exactly, we can round this value to the nearest hundredth:
2.231 kJ ≈ 2.23 kJ
Thus, the correct answer is not among the given options.